The position (x) of a particle of mass 1 kg moving along x-axis at time t is given by `(x=(1)/(2)t^(2))` meter. Find the work done by force acting on it in time interval from t=0 to t=3 s.

Explanation

`x=(1)/(2)t^(2)implies v=(dx)/(dt)=(1)/(2)(2t)=t`
`:. "At "t=0,v_(1)=0 implies "At "t=3s,v_(f)=3 ms^(-1)`
According to work-energy theorem,
`W=DeltaK=K_(f)-K_(i)`
`=(1)/(2)mv_(f)^(2)-(1)/(2)mv_(i)^(2)`
`=(1)/(2)xx1xx3^(2)=4.5 J`